Answer:
0.16joules
Explanation:
Using the relation for The gravitational potential energy
E= Mgh
Where,
E= Potential energy
h = Vertical Height
M = mass
g = Gravitational Field Strength
To find the vertical component of angle of launch Where the angle is 22°
h= sin theta
So E = mghsintheta
= 0.18 x 0.98 x 0.253 sin22
=0.16joules
Explanation:
Answer: D. The elements have the same number of valence electrons
Explanation: The chemical reactivity of elements is governed by the valence electrons present in the element.
The elements present in the same group or vertical column have similar valence configurations and thus behave similarly in chemical reactions or have similar bonding properties.
For Example: Both fluorine and chlorine belong to same family or group and both have 7 electrons in their valence shell and thus accept single electron to attain noble gas configuration.




thus both would bond with a cation bearing a single positive charge.
The magnitude and direction of the electric field in the wire are mathematically given as
![L &=[(v / L) v / m] \hat{i}](https://tex.z-dn.net/?f=L%20%26%3D%5B%28v%20%2F%20L%29%20v%20%2F%20m%5D%20%5Chat%7Bi%7D)
<h3>What is the magnitude and direction of the electric field in the wire?</h3>
Generally, the equation for is mathematically given as
A cylindrical wire that is straight and parallel to the x-axis has the following dimensions: length L, diameter d, resistivity p, diameter d, potential v, and z length. combining elements from both sides
E d 
![\begin{aligned}&-E \int_0^L d x=\int_v^0 d v \\\therefore E \cdot L &=v \\L &=[(v / L) v / m] \hat{i}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26-E%20%5Cint_0%5EL%20d%20x%3D%5Cint_v%5E0%20d%20v%20%5C%5C%5Ctherefore%20E%20%5Ccdot%20L%20%26%3Dv%20%5C%5CL%20%26%3D%5B%28v%20%2F%20L%29%20v%20%2F%20m%5D%20%5Chat%7Bi%7D%5Cend%7Baligned%7D)
In conclusion, the magnitude and direction of the electric field in the wire are given as
![L &=[(v / L) v / m]](https://tex.z-dn.net/?f=L%20%26%3D%5B%28v%20%2F%20L%29%20v%20%2F%20m%5D)
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A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.
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In the circuit outside of the battery the electrons have to expend all of their energy on the internal resistance of the battery which causes heating